A cube of metal is subjected to a hydrostatic pressure 4GPa . The...

B = - \dfrac{∆ P}{\dfrac{∆ V}{V}}$<br />$\dfrac{∆ V}{V} = - \dfrac{∆ P}{B}$<br />Magnitude is given by,<br />\left|\dfrac{∆ V}{V}\right| = \dfrac{∆ P}{B}$

Volume in terms of length can be represented as,

$3 \times \left|\dfrac{∆ L}{L}\right| = \dfrac{∆ P}{B}$

Therefore,

\left|\dfrac{∆ L}{L}\right| = \dfrac{1}{3} \times \dfrac{∆ P}{B}$</p><p>Substituting the values we get,<br />\left|\dfrac{∆ L}{L}\right| = \dfrac{1}{3} \times \dfrac{4 \times {10}^{9}}{8 \times {10}^{10}} = \dfrac{1}{60}$

Percantage change is,
$\dfrac{1}{60} \times 100 = \dfrac{5}{3} = 1 . 67$